B. Nastya Studies Informatics（数学）

B. Nastya Studies Informatics

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Today on Informatics class Nastya learned about GCD and LCM (see links below). Nastya is very intelligent, so she solved all the tasks momentarily and now suggests you to solve one of them as well.

We define a pair of integers $(a,b)$ good, if $GCD(a,b)=x$ and $LCM(a,b)=y$, where $GCD(a,b)$ denotes the greatest common divisor of $a$ and $b$, and $LCM(a,b)$ denotes the least common multiple of $a$ and $b$.

You are given two integers $x$ and $y$. You are to find the number of good pairs of integers $(a,b)$ such that $l\le a,b\le r$. Note that pairs $(a,b)$ and $(b,a)$ are considered different if $a\ne b$.

Input

The only line contains four integers $l,r,x,y$ ($1\le l\le r\le {10}^9$, $1\le x\le y\le {10}^9$).

Output

In the only line print the only integer — the answer for the problem.

Examples

input

Copy

1 2 1 2

output

Copy

2

input

Copy

1 12 1 12

output

Copy

4

input

Copy

50 100 3 30

output

Copy

0

Note

In the first example there are two suitable good pairs of integers $(a,b)$: $(1,2)$ and $(2,1)$.

In the second example there are four suitable good pairs of integers $(a,b)$: $(1,12)$, $(12,1)$, $(3,4)$ and $(4,3)$.

In the third example there are good pairs of integers, for example, $(3,30)$, but none of them fits the condition $l\le a,b\le r$.

题意：a,b为两个正整数，l<= a,b <=r。x=gcd（a，b），y=lcm（a，b）。gcd是a，b的最大公约数，lcm是a，b的最小公倍数。给出l，r，x，y，求有多少种可能的a，b。

思路：如果直接暴力，利用x*y=a*b 来做的话，肯定会T！现在，我们设 a=n*m,b=m*z.那么，x=m，y=n*m*z，a*b=n*m*m*z。如果我们来枚举a的所有可能的话，1e9，太大，所以我们考虑枚举n的所有可能！！我们先来确定n的范围，最小为1，n最大值为sqrt（n*z），确定了范围后，就可以开始写了。

   #include "iostream"
    using namespace std;
    int gcd(int a,int b)
    {
        if(b==0) return a;
        return gcd(b,a%b);
    }
    int main()
    {
        int l,r,x,y,i,num,ans=0;
        cin>>l>>r>>x>>y;
        num=y/x; if(y%x!=0){cout<<"0"<<endl;return 0;}//num=n*z
        for(i=1;i*i<=num;i++){
            int j=num/i;
            if(num%i==0&&gcd(i,j)==1&&l<=i*x&&i*x<=r&&l<=j*x&&j*x<=r) ans+=i==j?1:2;
        }
        cout<<ans<<endl;
    }